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vekshin1
3 years ago
13

Can someone help me with 5/8+2/5 please

Mathematics
2 answers:
Sedbober [7]3 years ago
7 0
To add fractions we have to find a common denominator and then add. To find the common denominator of the fractions we can multiply the denominators together to get 40.

Multiply 5/8 × 5/5 = 25/40
Multiply 2/5 × 8/8 = 16/40

Now add the fractions

25/40 + 16/40 = 41/40 or 1 1/40
marishachu [46]3 years ago
5 0
Well you would have to make the bottom number equal, so the lcm (least common multiple) of 5 and 8 is 40. Which means for the first you have to multiply 5 and 8 both by five to get 25/40. Then multiply 2 and 5 both by 8 so 16/40 then add. So 25/40 + 16/40 = 41/40.

So the answer is 41/40, which reduced ia 1 and 1/40.

Hope that helps!
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Sergio [31]

Answer:

The values of p in the equation are 0 and 6

Step-by-step explanation:

First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p

2

−7p−4=(2p+1)(p−4)

So then the equation looks like:

\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}

2p+1

p

−

(2p+1)(p−4)

2p

2

+5

=−

p−4

5

To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:

\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}

(2p+1)(p−4)

p

2

−4p

−

(2p+1)(p−4)

2p

2

+5

=−

(p−4)(2p+1)

10p+5

Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.

(p^2-4p)-(2p^2+5)=-(10p+5)(p

2

−4p)−(2p

2

+5)=−(10p+5)

Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p

2

−4p)−(2p

2

+5) first:

(p^2-4p)-(2p^2+5)=-p^2-4p-5(p

2

−4p)−(2p

2

+5)=−p

2

−4p−5

-p^2-4p-5=-10p+5−p

2

−4p−5=−10p+5

Combine like terms:

-p^2-4p+0=-10p−p

2

−4p+0=−10p

-p^2+6p=0−p

2

+6p=0

Factor:

p=0, p=6p

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Answer:

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